10.2  Solving Linear Equations in One Unknown

10.2.1 What a linear equation in one unknown is

Before we solve, we have to know what we're allowed to solve. A linear equation in one unknown is the simplest interesting kind of equation. Two words do all the work. One unknown means a single letter — usually x — and only that letter. Linear means the unknown appears only to the first power: just plain x, never x², never 1x, never √x.

Here is the cleanest way to think about it. No matter how messy an equation looks, if it's linear you can always tidy it — gather, combine, move everything to one side — until it reads in the standard form

ax + b = 0,   with a ≠ 0.

Here a is the number multiplying the unknown (its coefficient) and b is a plain number. We insist a ≠ 0 because if the x disappeared entirely, there would be nothing left to solve for.

Run a quick test on each one. Is there exactly one letter? Does it appear only as a plain first power? 3x − 5 = 1 passes. 2(x + 1) = x looks busier, but open the bag and it's still just first-power x's — it passes too. But x² = 4 has a square, 2x = 1 hides the unknown under a fraction bar, and √x = 3 has a root. Those three are not linear, and the routine in this lesson is not built for them (you'll meet squares in Stage 11).

10.2.2 Moving terms across the equals sign

Here is the single move that makes solving fast. Look at 3x + 5 = 11. We want the x-stuff alone, so the + 5 has to go. The honest, Lesson-10.1 way is to subtract 5 from both sides:

Notice the result: the + 5 vanished from the left and reappeared on the right as − 5. That is the whole trick. So we skip the middle line and just say it out loud:

A term may jump across the equals sign — but it flips its sign as it crosses.

This shortcut has a name: transposition, or "moving terms across." A + 5 becomes − 5; a − 8 becomes + 8; a + 2x on the right becomes − 2x on the left. It is never a new rule — it is always "add or subtract the same thing from both sides," just written in one step instead of two.

10.2.3 Combining like terms and making the coefficient one

Now we have everything we need for a full solve. When the unknown shows up on both sides, the plan is always the same three moves:

1. Send all the x-terms to one side and all the plain numbers to the other (each one flips its sign as it crosses). 2. Combine like terms on each side so you're left with a single x-term and a single number. 3. Divide both sides by the coefficient of x, so its coefficient becomes 1 and x stands alone.

Watch it on 5x − 3 = 2x + 9:

And we never trust an answer until we check it. Put x = 4 back into the original equation and make sure both pans really do match:

left: 5(4) − 3 = 20 − 3 = 17   ·   right: 2(4) + 9 = 8 + 9 = 17   ✓

Both sides equal 17, so x = 4 is right. That last "÷ 3" step is the amber move — dividing both sides by the coefficient — and it's almost always the very last thing you do.

10.2.4 Solving equations with parentheses

A pair of parentheses is a packed bag. Before you can move the things inside, you have to open the bag — and the tool for that is the distributive law from Stage 4: 2(x − 3) = 2·x − 2·3 = 2x − 6. Multiply the outside number into every term inside. Once the bag is open, you're back to the routine you already know.

Solve 2(x − 3) = x + 4:

Here the coefficient already came out to 1, so there was no dividing to do. Check in the original: left: 2(10 − 3) = 2·7 = 14; right: 10 + 4 = 14. ✓

10.2.5 Solving equations with fractions

Fractions look scary but they're a chore, not a mystery — and there's a one-shot trick to make them disappear. Multiply both sides by the least common multiple (LCM) of the denominators. That single move clears all the fractions at once, because the LCM is a multiple of every denominator (you cleared denominators just like this back in Stage 9).

Solve x+12 − x−13 = 1. The denominators are 2 and 3, so LCM(2, 3) = 6. Multiply every term by 6:

Why did 6 turn x+12 into 3(x+1)? Because 6 ÷ 2 = 3. Likewise 6 ÷ 3 = 2, and the lonely 1 on the right becomes 6 × 1 = 6. Now check in the original equation: 1+12 − 1−13 = 22 − 03 = 1 − 0 = 1. ✓

10.2.6 Checking solutions and special cases

You've seen me check every answer by substituting it back into the original equation. That's not a formality — it's the safety net that catches a flipped sign or a botched distribution. Always go back to the original, never to a tidied line (a tidied line could carry an early mistake forward).

Most linear equations have exactly one solution. But two surprising things can happen, and both show up when the x-terms cancel each other out. Then the equation collapses to a statement about plain numbers — and that statement is either always false or always true.

No solution

Solve 2x + 1 = 2x + 3. Move the 2x across and it cancels:

The x vanished and left 0 = 2, which is simply false. No number can make it true, so this equation has no solution. (And it makes sense: 2x + 1 is always exactly 2 less than 2x + 3 — they can never be equal.)

Infinitely many solutions (an identity)

Solve 2x + 2 = 2(x + 1). Open the bag on the right first:

The two sides are identical. Subtract one from the other and you get 0 = 0, which is always true. So every number is a solution — we call such an equation an identity.

Here's the quick way to tell the three cases apart for ax + b = cx + d: compare the x-coefficients first, then the numbers.